# Obtaining $\int_0^2\big(-\frac{x^2}4-5 \big)dx$ by a Riemann sum

EDIT: $R_n$ is the right endpoint of the Riemann sum

Calculate $R_n$ for the function $$f(x) = -\frac{x^2}4-5$$ on interval [0,2]

What I've done so far

$\Delta x = \frac 2n$ and $f(x_i^*) = \frac{2i}n$

Which leads to $$\sum_{i=1}^n f(x_i^*)\ \Delta x$$

And once you put the numbers in

$$\sum_{i=1}^n\left( -\frac{\left(\frac{2i}n\right)^2}4-5\right) \cdot \frac 2n$$

Which leads to my issue of expanding and getting an $i$ or $i^2$ alone.

• @T.Bongers Just added that in the question. Jan 17, 2016 at 21:43

You have $$\sum_{i=1}^n f(x_i^*)\ \Delta x=-\frac {2}{n^3} \sum_{i=1}^n i^2- \frac {10}{n}\sum_{i=1}^n 1$$ then use $$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$ which is proved here to get, as $n \to \infty$, $$\sum_{i=1}^n f(x_i^*)\ \Delta x \to -\frac{32}3$$

• Hey thanks dude! Appreciate that! Jan 17, 2016 at 22:03
• You are welcome. @impact_sv1 Jan 17, 2016 at 22:06